Formula for linear transformation of cross product Motivated by this question and by my own calculations where I try to find what is the curvature of a rigid transformation to my curve I get to the following equation ($\alpha$ is a curve, $\phi$ a linear transformation):

$\phi(\alpha') \times \phi(\alpha'') = \phi(\alpha' \times \alpha'')$

I wonder if it is true and conjecture that it is true because $\phi$ is an orthogonal linear transformation. So that the more general formula:

$Ma \times Mb = \det(M)(M^T)^{-1} (a \times b)$

holds. Now, what I ask you for is to prove this last formula (fix it if it is not true) or at least give a reference where I can encounter it. 
 A: The general formula is correct. Your original one holds only when $\phi$ is a proper (orientation-preserving) orthogonal linear map. What underlies all of this is (a) the way linear maps act on the exterior algebra:
$$(\Lambda^2M)(a\wedge b) = Ma\wedge Mb$$
and (b) the Hodge star isomorphism $\star\colon \Lambda^2\Bbb R^3 \to\Bbb R^3$. So the underlying action on the cross product of two vectors $a\times b$ is found by chasing around $\star (\Lambda^2M)(a\wedge b) = \star(\Lambda^2M)(\star(a\times b))$.
We don't need to be so fancy. The crucial thing is the "classical adjoint" formula 
$$M^{-1} = \frac1{\det M}(\text{cof }M)^\top,$$
where $\text{cof } M$ is the matrix of cofactors. Here is the typical computation: 
\begin{align*}
Me_1 \times Me_2 &= \sum m_{k1}e_k \times \sum m_{\ell 2}e_\ell = \sum_{k<\ell} \left|\begin{matrix} m_{k1} & m_{k 2} \\ m_{\ell 1} & m_{\ell 2}\end{matrix} \right| e_k\times e_\ell \\
&= \sum (\text{cof }M)_{i3}e_i = (\text{cof }M)e_3 = (\det M)(M^{-1})^\top e_3\\
&= (\det M)(M^{-1})^\top(e_1\times e_2).
\end{align*}
